Thank you for reporting, we will resolve it shortly
Q.
A particle of mass $m$ is acted upon by a force $F=t^{2}-kx$ . Initially, the particle is at rest at the origin. Then
NTA AbhyasNTA Abhyas 2020Oscillations
Solution:
$a=\frac{t^{2}}{m}-\frac{k}{m}x$
$\frac{d a}{d t}=\frac{1}{2}2t-\frac{k}{m}.\frac{d x}{d t}$
$\frac{d^{2} a}{d t^{2}}=\frac{2}{m}-\frac{k}{m}a$
$\frac{d^{2} a}{d t^{2}}+\frac{k}{m}a-\frac{2}{m}=0$
This is second order differential equation for acceleration, hence $a$ will be in simple harmonic.